Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x+2y &= 4 \\ 8x+2y &= 6\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = -2y+6$ Divide both sides by $8$ to isolate $x$ $x = {-\dfrac{1}{4}y + \dfrac{3}{4}}$ Substitute this expression for $x$ in the first equation. $-2({-\dfrac{1}{4}y + \dfrac{3}{4}}) + 2y = 4$ $\dfrac{1}{2}y - \dfrac{3}{2} + 2y = 4$ Simplify by combining terms, then solve for $y$ $\dfrac{5}{2}y - \dfrac{3}{2} = 4$ $\dfrac{5}{2}y = \dfrac{11}{2}$ $y = \dfrac{11}{5}$ Substitute $\dfrac{11}{5}$ for $y$ in the top equation. $-2x+2( \dfrac{11}{5}) = 4$ $-2x+\dfrac{22}{5} = 4$ $-2x = -\dfrac{2}{5}$ $x = \dfrac{1}{5}$ The solution is $\enspace x = \dfrac{1}{5}, \enspace y = \dfrac{11}{5}$.